Question: Let $R$ be the region enclosed by the line $y=2$ and the curve $2(x-1)^2$. $y$ $x$ ${y=2(x-1)^2}$ $ R$ $ 0$ $ 2$ $ 2$ A solid is generated by rotating $R$ about the line $y=2$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi\int^2_0[2-2(x-1)^2]\, dx$ (Choice B) B $\pi \int^2_0 2(x-1)^2\, dx$ (Choice C) C $\pi\int^2_0[2(x-1)^2]^2\, dx$ (Choice D) D $\pi\int^2_0[2-2(x-1)^2]^2\, dx$
Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=2(x-1)^2}$ Each slice is a cylinder. Let the width of each slice be $dx$ and let the radius of the base, as a function of $x$, be $r(x)$. Then, the volume of each slice is $\pi [r(x)]^2\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small width using a definite integral: $\int_a^b \pi [r(x)]^2\,dx$ This is called the disc method. What we now need is to figure out the expression of $r(x)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=2(x-1)^2}$ $ 0$ $ 2$ $ 2$ $r$ The radius is equal to the distance between the curve $y=2(x-1)^2$ and the line $y=2$. In other words, for any $x$ -value, this is the equation for $r(x)$ : $\begin{aligned} r(x)}&=2-2(x-1)^2} \end{aligned}$ Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(x)}]^2 \\\\ &=\pi\left[2-2(x-1)^2}\right]^2 \end{aligned}$ The leftmost endpoint of $R$ is at $x=0$ and the rightmost endpoint is at $x=2$. So the interval of integration is $[0,2]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^2 \left[\pi(2-2(x-1)^2)\right]^2dx \\\\ &=\pi\int_0^2 \left[2-2(x-1)^2)\right]^2dx \end{aligned}$